推 llww :厲害,謝謝您的詳細說明 05/15 18:40
※ 引述《llww (開心渡過每一天)》之銘言：
: 請問各位先進如下的問題:
: 設三角形ABE在正方形ABCD的外側，AE = BE；若F在線段AE上，且
: EF = AB，BF = BD，試証 角AEB =(180/7) 度.
: 謝謝各位
角EAB = k
角AEB = p
AB = a
AF = b
a^2 + b^2 - 2abcos(k) = 2a^2
(a + b)^2 + a^2 - 2a(a + b)cos(k) = (a + b)^2
=> a = 2(a + b)cos(k)
=> b = a[1 - 2cos(k)]/[2cos(k)]
且cos(k) < 1/2 => k > π/3
2a^2 = a^2 + b^2 + 2ab - 2ab[1 + cos(k)]
=> 2a^2 = a^2/[4(cos(k))^2] - a^2[1 - 2cos(k)][1 + cos(k)]/cos(k)
令x = cos(k)
=> 8x^2 = 1 - 4x[1 - 2x][1 + x]
=> 8x^3 - 4x^2 - 4x + 1 = 0
=> cos(π/7)是滿足x的最大根 但顯然不是其解
所以次大的根cos(3π/7)才是解
cos(k) = cos((π-p)/2)
=> p = π - 2k = π/7
所以角AEB =(180/7) 度
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